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利用因式分解简便计算 2004^2-4008×2005+2005^2 9.9^2+9.9×0.2+0.01 2003^2-2001^2分之1001
已知4m+n=90,2m-3n=10,求(m+2n)^2-(3m-n)^的值
(a^2+4)^2-16a^2
-x^2-4y^2+4xy
a^2(x-y)-b^2(x-y)
人气:167 ℃ 时间:2020-04-09 10:41:05
解答
2004^2-4008×2005+2005^2
=2004^2-2×2004×2005+2005^2
=(2004-2005)²
=(-1)²
=1
9.9^2+9.9×0.2+0.01
=9.9^2+2×9.9×0.1+0.1 ²
=(9.9+0.1)²
=10²
=100
2003^2-2001^2分之1001
=(2003+2001)(2003-2001)分之1001
=4004×2分之1001
=8分之1
(m+2n)^2-(3m-n)^2
=(m+2n+3m-n)(m+2n-3m+n)
=(4m+n)(-2m+3n)
=90×(-10)
=-900
(a^2+4)^2-16a^2
=(a²+4+4a)(a²+4-4a)
=(a+2)²(a-2)²
-x^2-4y^2+4xy
=-(x²-4xy+4y²)
=-(x-2y)²
a^2(x-y)-b^2(x-y)
=(a²-b²)(x-y)
=(a+b)(a-b)(x-y)
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